Respiratory alkalosis (acute/uncompensated)problem #1 explanation: ph 7.56, paco2 20, hco3 20, pao2 88 although the technical classification is a partially compensated respiratory alkalosis, the correct final interpretation is an acute or uncompensated respiratory alkalosis. steps to interpretation step 1. check ph is it acidemia or alkalemia? answer: alkalemiareasoning: normal ph range is 7.35-7.45 greater than 7.45 = alkalemia step 2. check paco2 is respiratory the primary cause? answer: yes reasoning: normal paco2 = 35-45 mmhg. paco2 is < 35 (abnormal)if paco2 is abnormal and goes in the opposite direction of ph, then it is a respiratory cause.again think of co2 as an acid ( co2 + h2o h+ + hco3) = ph hco3 = ph paco2 step 3. check hco3 is metabolic the primary cause? answer: no reasoning: normal hco3 = 22-26 meq/l hco3 is 20 and outside of normal range. although hco3 is abnormal, it is going in the opposite direction of ph (hco3; ph)if hco3 is abnormal and goes in the same direction as ph, then it is a metabolic cause. ( hco3 ph; hco3 ph) step 4. is the body compensating?answer: perhaps. calculation in step 6 is required to determine if the change in hco3 is due to hydrolysis or compensation.reasoning: hco3 is less than normal, so either the kidney is compensating (lowering hco3 in order to lower ph) or it is due to the hydrolysis reaction. step 5: determine the technical classification reasoning: ph, paco2, hco3 answer: technical classification is either uncompensated respiratory alkalosis or partially compensated respiratory alkalosis (see page 12 of oakes abg pocket guide)functional classification is either acute orchronic respiratory alkalosis (see page 14 of oakes abg pocket guide).again, calculations are required to distinguish between the two possibilities. step 6. determine if compensation is appropriate or are there other primary causes respiratory is the primary cause:paco2 and ph relationship:paco2 has 20 mmhg, ph has 0.16.this is consistent with an acute change due to hydrolysis (every 10 mmhg paco2 0.08 ph) (therefore, 2 x 0.08 = 0.16) (7.40 + 0.16 = 7.56) paco2 and hco3 relationship:paco2 has 20mmhg, hco3 has 4 meq this is consistent with an acute change due to hydrolysis (every 10 mmhg paco2 2 meq hco3) (therefore, 2 x 2 = 4) (24 4 = 20)this is not considered compensation. answer: where the decrease in hco3 is due to hydrolysis, and not compensation by the kidneys, compensation (none, in this example) is appropriate and consistent with acute (uncompensated) respiratory alkalosis (and not partially compensated respiratory alkalosis).there is no evidence of other primary causes (more on this later). steps 7 - 10 these will be considered later as we move on past the introductory basics. step 11. final interpretationacute (uncompensated) respiratory alkalosis summary: as mentioned above, calculations are always required to confirm whether the technical classification is the same as the final interpretation. here is an instance where the technical classification differs from the final interpretation.without calculating for hydrolysis, reliance on only a technical classification (and/or the functional classification) in this situation, would lead one to erroneously conclude that the kidney is partly compensating, when it is not. note: technical classification is taught (and needed) as an organizational starting point and functional classification is the common terminology used for technical classification, especially among physicians. remember: classification does not always equal interpretation!
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